1. C# / Говнокод #25330

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    using System;
using var = innerx.var_;
 
namespace innerx {    
    public class var___ : var
    {
        public var var => new var();
    }    
    public class var__ : var
    {
        public var___ var() => new var___();
    }
    public class var_ {
        public static readonly var__ var = new var__();
        public override string ToString() => "var";
    }
}
 
public class foo {
    public static void Main() {
        var var = var.var.var().var;
        Console.WriteLine(var);
    }
}

    https://ideone.com/2N2MOU

    Запостил: cykablyad, 27 Января 2019

    Комментарии (8) RSS

    • ava 1024-- 27.01.2019 16:03 # +4

      Варварство!
      Ответить

    • ava JIyHHblu_nemyx 27.01.2019 16:41 # −1

      Два решения на "PHP". Первое: 
      <?php

class var___ extends var_ {
    public function __get ($name) {if($name === 'var') return new var_();}
    public function __isset ($name) {return $name === 'var';}
}    

class var__ extends var_ {
    public function __get ($name) {if($name === 'var') return function() {return new var___();};}
    public function __isset ($name) {return $name === 'var';}
}

class var_ {
    public function __get ($name) {if($name === 'var') return new var__();}
    public function __isset ($name) {return $name === 'var';}
    public function __toString() {return 'var';}
}

$var = new var_();
$var = $var->var->var; 
$var = $var()->var;

print($var);


      Второе: 
      <?php

class var___ extends var_ {
    public function __get ($name) {if($name === 'var') return new var_();}
    public function __isset ($name) {return $name === 'var';}
    public function __invoke() {return $this;}
}    

class var__ extends var_ {
    public function __get ($name) {if($name === 'var') return new var___();}
    public function __isset ($name) {return $name === 'var';}
}

class var_ {
    public function __get ($name) {if($name === 'var') return new var__();}
    public function __isset ($name) {return $name === 'var';}
    public function __toString() {return 'var';}
}

$var = new var_();
$var = $var->var->var;
$var = $var()->var;

print($var);
      Ответить

    • ava Elvenfighter 29.01.2019 12:46 # +1

      Ну, такое
      struct derived {
   explicit derived() = default;
   int override() const { return 0; }
};

using override = derived;

int main()
{
    override override;
    return override.override();
}
      Ответить

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